Calculate $\int_{D}||\nabla f||^2 dA$, where $f(x,y)=y-x^2+1$, and $D$ is this region

Question

Calculate $$\int_{D}||\nabla f||^2 dA$$ where $f(x,y)=y-x^2+1$, and $D$ is this region $\{(x,y)|f(x,y)\geq 0, y \leq 0\}$

$$\nabla f=(-2x,1)\to ||\nabla f||^2=4x^2+1$$

I have found the function, but not able to understand what the limits of integration would be.

If we have that $y\leq 0$, then $-\infty \lt y \leq 0$.

If $y=0$, then $f(x,y)=-x^2+1\geq 0\to x^2\leq 1\to x\in[-1,1]$

But let's say $y$ = -4, then $f(x,y)=-3-x^2\geq 0 \to 3+y^2 \geq 0\to x\in[-\infty,\infty)$

Taking the intersection $\color{red}{\text{ (union ?) }}$, we see that $x\in [-1,\infty]$

So the integral would become:

$$\int_{-\infty}^{0}\int_{-1}^{\infty}4x^2+1\text { dx dy}$$

Is this correct?

Answer 1

The first step is always to draw a picture. First we have

$$ f(x,y) \ge 0 \rightarrow y-x^2+1\ge 0 \rightarrow y \ge x^2-1 $$

Combined with $y \le 0 $, the region looks like this

I'm sure you can figure out the integral limits from here. Hint: The $y$ limits should depend on $x$

Answer 2

If $y = -4$ then $-3-x^2 \ge 0$ then $-3 \geq x^2 \ge 0$.

Hence no such $x$ exists.

$y$ doesn't really take value from $-\infty $ to $0$. There are values from $(-\infty, 0]$ that $y$ doesn't take, for example $y=-4$ as shown earlier.

From $y-x^2+1 \geq 0$, we have $y \geq x^2-1 \geq -1$.

Remark:

The limit for $x$ is not quite right too. Try to sketch out the region, plot out the curve $y=x^2-1$, and studying the region might help. Think of after you fix $y$, what are the values that $x$ can take.

Answer 3

Once we sketch the region we see that we are integrating slices in the $y$ direction bounded below by the curve $y=x^2-1$ and above by $0$. We sum these slices from $x=-1$ to $x=1$.

Your integrand is $$ \nabla f\cdot \nabla f=4x^2+1 $$ yielding the integral $$ \int_{-1}^1\int_{x^2-1}^0 (4x^2+1)\mathrm dy\mathrm dx $$